JEE PYQ: Motion in a Plane - Question ID 8cc83c23d8e8 (JEE Main 2023)

ID: 8cc83c23d8e8JEE Main 2023Single Correct MCQ

Two objects are projected with same velocity 'u' however at different angles α\alpha and β\beta with the horizontal. If α+β=90∘\alpha+\beta=90^\circ, the ratio of horizontal range of the first object to the 2nd object will be :

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Step-by-step Explanation

Core Formula & Concept:

In projectile motion, the horizontal range (RR) of an object projected with initial velocity uu at an angle θ\theta with the horizontal is given by: R=u2sin⁡(2θ)gR = \frac{u^2 \sin(2\theta)}{g} where:

  • uu is the initial velocity,
  • θ\theta is the projection angle,
  • gg is the acceleration due to gravity.
This formula arises from the fact that the horizontal range depends on the horizontal component of velocity (ucos⁡θu \cos \theta) and the total time of flight (T=2usin⁡θgT = \frac{2u \sin \theta}{g}). The product of these two quantities yields the range formula above.

The question involves two objects projected with the same speed uu but at angles α\alpha and β\beta, where α+β=90∘\alpha + \beta = 90^\circ. We are to find the ratio of their horizontal ranges.

Step-by-Step Derivation:

Step 1: Write the range expressions for both objects.
For the first object (angle α\alpha): R1=u2sin⁡(2α)gR_1 = \frac{u^2 \sin(2\alpha)}{g} For the second object (angle β\beta): R2=u2sin⁡(2β)gR_2 = \frac{u^2 \sin(2\beta)}{g}

Step 2: Use the given condition α+β=90∘\alpha + \beta = 90^\circ.
Since α+β=90∘\alpha + \beta = 90^\circ, we can express β\beta as: β=90∘−α\beta = 90^\circ - \alpha

Step 3: Substitute β\beta into R2R_2.
Now, compute sin⁡(2β)\sin(2\beta): sin⁡(2β)=sin⁡(2(90∘−α))=sin⁡(180∘−2α)=sin⁡(2α)\sin(2\beta) = \sin(2(90^\circ - \alpha)) = \sin(180^\circ - 2\alpha) = \sin(2\alpha) This follows from the trigonometric identity: sin⁡(180∘−x)=sin⁡x\sin(180^\circ - x) = \sin x.

Step 4: Compare R1R_1 and R2R_2.
Substituting back: R2=u2sin⁡(2α)g=R1R_2 = \frac{u^2 \sin(2\alpha)}{g} = R_1 Thus, the ranges are equal: R1:R2=1:1R_1 : R_2 = 1 : 1

Common Traps & Exam Tip:

Trap 1: Misapplying the range formula.
Some students mistakenly use R=u2sin⁡θgR = \frac{u^2 \sin \theta}{g} instead of R=u2sin⁡(2θ)gR = \frac{u^2 \sin(2\theta)}{g}. This leads to incorrect results. Always double-check the formula for range. Trap 2: Overlooking the trigonometric identity.
Students may fail to recognize that sin⁡(2β)=sin⁡(2α)\sin(2\beta) = \sin(2\alpha) when α+β=90∘\alpha + \beta = 90^\circ. This step is crucial for simplifying the ratio. Remember that sin⁡(180∘−x)=sin⁡x\sin(180^\circ - x) = \sin x. Trap 3: Assuming unequal ranges due to different angles.
Intuitively, one might think that different angles lead to different ranges. However, the condition α+β=90∘\alpha + \beta = 90^\circ ensures that the ranges are equal, as shown in the derivation. Always rely on the math rather than intuition alone. Exam Tip:
When dealing with projectile motion problems involving complementary angles (i.e., α+β=90∘\alpha + \beta = 90^\circ), remember that the ranges will be equal. This is a common pattern in JEE problems and can save time during the exam.